A car is moving in a circular horizontal track of radius 10 m with a constant speed of 10 m/s. A plu

Q: Wait, why does the bob swing outwards in the first place? Is there a force pushing it out?

From the ground frame (inertial frame), there is NO outward force! The bob actually wants to go straight due to inertia. The car is turning inwards, so the rod has to pull the bob inwards (via $T \sin\theta$) to force it to turn with the car. The 'outward push' you feel is just your body's inertia resisting the turn. If you solve this from inside the car (a non-inertial frame), you would invent a 'pseudo force' (centrifugal force) pushing it out to make the math work.

Q: Why didn't they give us the mass of the plumb bob? Can we solve it without it?

Yes! This is a beautiful feature of Newtonian mechanics. The force required to turn the bob (centripetal force) is proportional to its mass ($m$). But the force holding it down (gravity) is also proportional to its mass ($m$). When you divide the two equations, the mass cancels out completely. A feather and a bowling ball would hang at the exact same angle!

Q: Why do we use $T \sin\theta$ for the centripetal force and not $T \cos\theta$?

It depends entirely on where the angle $\theta$ is measured from. The problem states $\theta$ is the angle with the **vertical**. If you draw a right triangle, the component adjacent to the vertical is $T \cos\theta$ (which points up), and the component opposite to the vertical is $T \sin\theta$ (which points horizontally towards the center of the circle).

Q: What would happen if the car started moving faster?

Look at our master equation: $\tan\theta = \frac{v^2}{Rg}$. If the velocity $v$ increases, the numerator increases, which means $\tan\theta$ increases. Since $\tan\theta$ grows as $\theta$ grows (between $0$ and $90^\circ$), the angle $\theta$ would increase. The bob would swing further outwards!

Q: Can I solve this using the Centrifugal force concept?

Absolutely! If you sit inside the car (a rotating, non-inertial frame), the bob appears to be at rest. You must apply an outward centrifugal pseudo-force of magnitude $\frac{mv^2}{R}$. Now, balance the forces: Upward $T \cos\theta$ balances downward $mg$, and inward $T \sin\theta$ balances outward $\frac{mv^2}{R}$. You get the exact same equations!

Question

A car is moving in a circular horizontal track of radius $10 	ext{ m}$ with a constant speed of $10 	ext{ m/s}$. A plumb bob is suspended from the roof of the car by a light rigid rod. The angle made by the rod with the vertical is (Take $g = 10 	ext{ m/s}^2$)

Options:

(A) zero

(B) $30^\circ$

(C) $45^\circ$

(D) $60^\circ$

Sigma Percentile · Accepted Answer

### The Thrill of the Turn

Imagine you are sitting in the passenger seat of a car. The driver suddenly takes a sharp, high-speed turn along a circular track. What happens to you? You feel an invisible hand pushing you outwards, pressing you against the car door. This everyday experience is the perfect gateway to understanding the physics of circular motion and non-inertial frames.

In our problem, instead of a passenger, we have a simple plumb bob suspended from the roof of the car by a light rigid rod. As the car moves in a horizontal circle of radius $R = 10 	ext{ m}$ with a constant speed of $v = 10 	ext{ m/s}$, the bob doesn't just hang straight down. It swings outwards, making a mysterious angle $	heta$ with the vertical. Our mission is to decode this angle.

### Analyzing the Setup: The Ground Frame Perspective

To solve this elegantly, let's step outside the car and observe the situation from the ground—an inertial frame of reference. From here, we see the car moving in a circle, and we see the plumb bob also moving in that exact same horizontal circle.

For any object to move in a circle, the universe demands a toll: the **Centripetal Force**. This force must always point towards the center of the circle. But who is paying this toll for our plumb bob? To find out, we must draw a Free Body Diagram (FBD).

Only two real forces are acting on the bob:
1. The gravitational pull of the Earth, acting straight down: $mg$.
2. The tension in the rigid rod, pulling along its length: $T$.

### The Master Equations

Since the tension $T$ is acting at an angle $	heta$ to the vertical, it's doing two jobs at once. We can resolve it into two perpendicular components to see this clearly.

The vertical component is $T \cos	heta$. Since the bob is neither flying up through the roof nor crashing through the floor, its vertical acceleration is zero. This gives us our first master equation—the equation of vertical equilibrium:

$$T \cos	heta = mg$$

The horizontal component is $T \sin	heta$. This component points directly towards the center of the circular track. This is our hero! It is the force providing the necessary centripetal acceleration ($a_c = \frac{v^2}{R}$). This gives us our second master equation— Newton's Second Law in the radial direction:

$$T \sin	heta = \frac{mv^2}{R}$$

### The Elegant Cancellation

We now have a system of two equations. We want to find $	heta$, but we don't know the tension $T$, and we don't even know the mass $m$ of the bob!

This is where the magic of algebra steps in. If we divide the horizontal equation by the vertical equation, something beautiful happens:

$$\frac{T \sin	heta}{T \cos	heta} = \frac{\frac{mv^2}{R}}{mg}$$

The unknown tension $T$ cancels out. The unknown mass $m$ cancels out. We are left with a pure, elegant kinematic relationship that governs all such banking and pendulum problems:

$$	an	heta = \frac{v^2}{Rg}$$

### Final Calculation

Now, it's just a matter of plugging in the numbers provided in the problem. We are given $v = 10 	ext{ m/s}$, $R = 10 	ext{ m}$, and we are told to take $g = 10 	ext{ m/s}^2$.

$$	an	heta = \frac{(10)^2}{(10)(10)}$$

$$	an	heta = \frac{100}{100} = 1$$

We know from basic trigonometry that the angle whose tangent is $1$ is $45^\circ$.

$$	heta = 45^\circ$$

And there we have it! The rod will make an angle of $45^\circ$ with the vertical. The beauty of this result is its universality—whether you hang a tiny pebble or a massive bowling ball, as long as the speed and radius are the same, the angle will always be exactly $45^\circ$.

Animated Solution for Physics - Laws of Motion: A car is moving in a circular horizontal track of radius $10 m$ with a constant speed of $10 m/s$ . A plumb bob is suspended from the roof of the car by a light rigid rod. The angle made by the rod with the vertical is (Take $g = 10 m/s^{2}$ )

Select Answer:

Visualized Solution

Visualizing the Physical Setup

Identifying the Forces

Resolving the Tension

Vertical Equilibrium

Horizontal Dynamics

Eliminating Tension $T$

Simplifying the Equation

Substituting the Values

Final Calculation

The Sigma Insight: Dynamics of Circular Motion

The Thrill of the Turn

Analyzing the Setup

The Master Equations

The Elegant Cancellation

Final Calculation

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Select Answer:

Visualized Solution

Visualizing the Physical Setup

Identifying the Forces

Resolving the Tension

Vertical Equilibrium

Horizontal Dynamics

Eliminating Tension .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T

Simplifying the Equation

Substituting the Values

Final Calculation

The Sigma Insight: Dynamics of Circular Motion

The Thrill of the Turn

Analyzing the Setup

The Master Equations

The Elegant Cancellation

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The minimum velocity (in ms−1) with which a car driver must traverse a flat curve of radius 150 m and coefficient of friction 0.6 to avoid skidding is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A bead of mass m stays at point P(a,b) on a wire bent in the shape of a parabola y=4Cx2 and rotating with angular speed ω (see figure). The value of ω is (neglect friction)

Eliminating Tension $T$

The minimum velocity (in $ms^{- 1}$ ) with which a car driver must traverse a flat curve of radius $150 m$ and coefficient of friction $0.6$ to avoid skidding is

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